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 MFP August 20th, 2013 04:26 AM

Question regarding a critical point formula

Hey everyone,

I fear that on my final I'll be given an equation that ressembles $ax^3 + bx^2 + cx + d$, and I won't be able to find the critical points.

I know that when you have the form $ax^2 + bx + c$, you can use $(-b +/- sqrt(b^2-4ac))/(2a)$.

Is there such a method for $ax^3 + bx^2 + cx + d$ ?

 agentredlum August 20th, 2013 04:58 AM

Re: Question regarding a critical point formula

There is a cubic formula

http://en.m.wikipedia.org/wiki/Cubic_function

But it is a bit complicated to use so even if you get a cubic function it is better to use the techniques of precalculus to determine rational roots (if any).

:D

 MFP August 20th, 2013 08:47 AM

Re: Question regarding a critical point formula

ouch, I don't think I'll be able to memorize that formula in time haha. Alright, I'll use other methods. Thanks agentredlum!

 John Marsh August 21st, 2013 04:19 AM

Re: Question regarding a critical point formula

"Solution: Given ax³+bx²+cx+d
let f(x)= ax³+bx²+cx+d
f'(x)=3ax²+2bx+c,where f'(x)=0
then there are two critical point ,and at least two real roots
there for we can use the formula x=[-2b±v(4b²-12ac)]/6a"

 greg1313 August 21st, 2013 09:13 AM

Re: Question regarding a critical point formula

Quote:
 Originally Posted by John Marsh then there are two critical point ,and at least two real roots
Not true. Consider x³ + x² + 5x + 3, for example.

 HallsofIvy August 21st, 2013 09:48 AM

Re: Question regarding a critical point formula

Quote:
 Originally Posted by John Marsh "Solution: Given ax³+bx²+cx+d let f(x)= ax³+bx²+cx+d f'(x)=3ax²+2bx+c,where f'(x)=0 then there are two critical point ,and at least two real roots there for we can use the formula x=[-2b±v(4b²-12ac)]/6a"
I don't believe that was the question- although I thought it was at first, myself.

IF the original function were a cubic, then the derivative would be a quadratic but the OP ways he/she solved a quadratic before.
I would interpret this as asking what to do if the derivative itself is a cubic polynomial.

MFP, I doubt that you would be expected to know the cubic formula. IF the derivative turns out to be cubic, try simple integer values for x- bet on it being easily factorable.

 slipjack August 21st, 2013 02:42 PM

Let c = p - q. Assume a ? 0, b, p and q are real numbers.
However if we have ax² + bx + p = q then we have ax² + bx + p - q = 0.
The typical quadratic formula would be $x\,=\,\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2a}$.
To prove the quadratic formula is quite easy e.g. divide both sides by a then . . .
The quadratic formula is true if and only if ax² + bx + c = 0.

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